{"id":1161,"date":"2021-11-11T17:37:25","date_gmt":"2021-11-11T17:37:25","guid":{"rendered":"https:\/\/courses.lumenlearning.com\/calculus3\/chapter\/defining-polar-coordinates\/"},"modified":"2022-10-20T23:25:36","modified_gmt":"2022-10-20T23:25:36","slug":"defining-polar-coordinates","status":"publish","type":"chapter","link":"https:\/\/courses.lumenlearning.com\/calculus3\/chapter\/defining-polar-coordinates\/","title":{"raw":"Defining Polar Coordinates","rendered":"Defining Polar Coordinates"},"content":{"raw":"<div class=\"textbox learning-objectives\" data-type=\"abstract\">\r\n<h3>Learning Outcomes<\/h3>\r\n<ul>\r\n \t<li>Locate points in a plane by using polar coordinates<\/li>\r\n \t<li>Convert points between rectangular and polar coordinates<\/li>\r\n<\/ul>\r\n<\/div>\r\n<section id=\"fs-id1167794030048\" data-depth=\"1\">\r\n<p id=\"fs-id1167794039551\">To find the coordinates of a point in the polar coordinate system, consider Figure 1. The point [latex]P[\/latex] has Cartesian coordinates [latex]\\left(x,y\\right)[\/latex]. The line segment connecting the origin to the point [latex]P[\/latex] measures the distance from the origin to [latex]P[\/latex] and has length [latex]r[\/latex]. The angle between the positive [latex]x[\/latex] -axis and the line segment has measure [latex]\\theta [\/latex]. This observation suggests a natural correspondence between the coordinate pair [latex]\\left(x,y\\right)[\/latex] and the values [latex]r[\/latex] and [latex]\\theta [\/latex]. This correspondence is the basis of the <strong>polar coordinate system<\/strong>. Note that every point in the Cartesian plane has two values (hence the term <em data-effect=\"italics\">ordered pair<\/em>) associated with it. In the polar coordinate system, each point also has two values associated with it: [latex]r[\/latex] and [latex]\\theta [\/latex].<\/p>\r\n\r\n<figure id=\"CNX_Calc_Figure_11_03_001\"><figcaption><\/figcaption>\r\n\r\n[caption id=\"\" align=\"aligncenter\" width=\"304\"]<img src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/4175\/2019\/04\/11234807\/CNX_Calc_Figure_11_03_001.jpg\" alt=\"A point P(x, y) is given in the first quadrant with lines drawn to indicate its x and y values. There is a line from the origin to P(x, y) marked r and this line make an angle \u03b8 with the x axis.\" width=\"304\" height=\"309\" data-media-type=\"image\/jpeg\" \/> Figure 1. An arbitrary point in the Cartesian plane.[\/caption]<\/figure>\r\nAs we can observe above, the right triangle pictured above implies a set of relationships between [latex] x, y, r, \\:\\text{and}\\: \\theta [\/latex].\u00a0 We first recall those relationships below.\r\n<div class=\"textbox examples\">\r\n<h3>Recall: Right Triangle Trigonometry<\/h3>\r\nGiven a right triangle with an acute angle of [latex]\\theta[\/latex],\r\n<p style=\"text-align: left; padding-left: 60px;\">[latex]\\begin{align}&amp;\\sin \\left(\\theta \\right)=\\frac{\\text{opposite}}{\\text{hypotenuse}} \\\\ &amp;\\cos \\left(\\theta \\right)=\\frac{\\text{adjacent}}{\\text{hypotenuse}} \\\\ &amp;\\tan \\left(\\theta \\right)=\\frac{\\text{opposite}}{\\text{adjacent}} \\end{align}[\/latex]<\/p>\r\nA common mnemonic for remembering these relationships is SohCahToa, formed from the first letters of \"Sine is opposite over hypotenuse, Cosine is adjacent over hypotenuse, Tangent is opposite over adjacent.\"\r\n\r\nThe side lengths of the right triangle with legs [latex] a [\/latex] and [latex] b [\/latex] and hypotenuse\u00a0[latex] c [\/latex] are related through the Pythagorean Theorem:\u00a0[latex] a^2 + b^2 = c^2 [\/latex]\r\n\r\n<\/div>\r\n<p id=\"fs-id1167794070618\">Using right-triangle trigonometry, the following equations are true for the point [latex]P\\text{:}[\/latex]<\/p>\r\n\r\n<div id=\"fs-id1167794187190\" class=\"unnumbered\" style=\"text-align: center;\" data-type=\"equation\" data-label=\"\">[latex]\\cos\\theta =\\frac{x}{r}\\:\\:\\text{so}\\:\\:x=r\\cos\\theta [\/latex]<\/div>\r\n&nbsp;\r\n<div id=\"fs-id1167794170718\" class=\"unnumbered\" style=\"text-align: center;\" data-type=\"equation\" data-label=\"\">[latex]\\sin\\theta =\\frac{y}{r}\\:\\:\\text{ so }\\:\\:y=r\\sin\\theta [\/latex].<\/div>\r\n&nbsp;\r\n<p id=\"fs-id1167793785830\">Furthermore,<\/p>\r\n\r\n<div id=\"fs-id1167793821731\" class=\"unnumbered\" style=\"text-align: center;\" data-type=\"equation\" data-label=\"\">[latex]{r}^{2}={x}^{2}+{y}^{2}\\text{ and }\\tan\\theta =\\frac{y}{x}[\/latex].<\/div>\r\n&nbsp;\r\n<p id=\"fs-id1167794056124\">Each point [latex]\\left(x,y\\right)[\/latex] in the Cartesian coordinate system can therefore be represented as an ordered pair [latex]\\left(r,\\theta \\right)[\/latex] in the polar coordinate system. The first coordinate is called the <span data-type=\"term\">radial coordinate<\/span> and the second coordinate is called the <span data-type=\"term\">angular coordinate<\/span>. Every point in the plane can be represented in this form.<\/p>\r\n<p id=\"fs-id1167794170727\">Note that the equation [latex]\\tan\\theta =\\frac{y}{x}[\/latex] has an infinite number of solutions for any ordered pair [latex]\\left(x,y\\right)[\/latex]. However, if we restrict the solutions to values between [latex]0[\/latex] and [latex]2\\pi [\/latex] then we can assign a unique solution to the quadrant in which the original point [latex]\\left(x,y\\right)[\/latex] is located. Then the corresponding value of <em data-effect=\"italics\">r<\/em> is positive, so [latex]{r}^{2}={x}^{2}+{y}^{2}[\/latex].<\/p>\r\n\r\n<div id=\"fs-id1167793948751\" class=\"theorem\" data-type=\"note\">\r\n<div data-type=\"title\">\r\n<div class=\"textbox shaded\">\r\n<h3 style=\"text-align: center;\" data-type=\"title\">theorem: Converting Points between Coordinate Systems<\/h3>\r\n\r\n<hr \/>\r\n<p id=\"fs-id1167794024884\">Given a point [latex]P[\/latex] in the plane with Cartesian coordinates [latex]\\left(x,y\\right)[\/latex] and polar coordinates [latex]\\left(r,\\theta \\right)[\/latex], the following conversion formulas hold true:<\/p>\r\n\r\n<div id=\"fs-id1167794066096\" style=\"text-align: center;\" data-type=\"equation\">[latex]x=r\\cos\\theta \\:\\text{and}\\:y=r\\sin\\theta [\/latex],<\/div>\r\n&nbsp;\r\n<div id=\"fs-id1167793834620\" style=\"text-align: center;\" data-type=\"equation\">[latex]{r}^{2}={x}^{2}+{y}^{2}\\:\\text{and}\\:\\tan\\theta =\\frac{y}{x}[\/latex].<\/div>\r\n&nbsp;\r\n<p id=\"fs-id1167793913264\">These formulas can be used to convert from rectangular to polar or from polar to rectangular coordinates.<\/p>\r\n\r\n<\/div>\r\n<\/div>\r\n<\/div>\r\nAs we can note carefully above, the equation [latex]\\tan\\theta =\\frac{y}{x}[\/latex] is <em>not<\/em> expressed using the inverse tangent function.\u00a0 The reason for this is because of how the domain restriction of the tangent function leads to a restricted range of the inverse tangent function, reviewed below.\r\n<div class=\"textbox examples\">\r\n<h3>Recall: Using The Inverse Tangent Function in the coordinate plane<\/h3>\r\nIf [latex] -\\frac{\\pi}{2} &lt; \\theta &lt; \\frac{\\pi}{2} [\/latex], and [latex] \\tan \\theta = \\frac{y}{x} [\/latex], then [latex] \\theta = \\tan^{-1} \\left( \\frac{y}{x} \\right) [\/latex]\r\n<p style=\"padding-left: 30px;\">That is, the inverse tangent function has a range of [latex] \\left(-\\frac{\\pi}{2}, \\frac{\\pi}{2}\\right) [\/latex], meaning that it always produces a positive angle in Quadrant I or a negative angle in Quadrant IV.<\/p>\r\nIf [latex] \\frac{\\pi}{2} &lt; \\theta &lt; \\frac{3\\pi}{2} [\/latex] and [latex] \\tan \\theta = \\frac{y}{x} [\/latex], then [latex] \\theta = \\tan^{-1} \\left( \\frac{y}{x} \\right) + \\pi [\/latex]\r\n<p style=\"padding-left: 30px;\">In other words, if the point [latex] \\left(x, y \\right)[\/latex] is in Quadrant II or III, the preceding rule means that you must add [latex] \\pi [\/latex] to the output of the inverse tangent function to produce an angle in the correct quadrant.<\/p>\r\n\r\n<\/div>\r\n<div id=\"fs-id1167794030722\" data-type=\"example\">\r\n<div id=\"fs-id1167794003809\" data-type=\"exercise\">\r\n<div id=\"fs-id1167794044637\" data-type=\"problem\">\r\n<div data-type=\"title\">\r\n<div class=\"textbox exercises\">\r\n<h3>Example: Converting between Rectangular and Polar Coordinates<\/h3>\r\n<div id=\"fs-id1167794044637\" data-type=\"problem\">\r\n<p id=\"fs-id1167793814959\">Convert each of the following points into polar coordinates.<\/p>\r\n\r\n<ol id=\"fs-id1167793906173\" type=\"a\">\r\n \t<li>[latex]\\left(1,1\\right)[\/latex]<\/li>\r\n \t<li>[latex]\\left(-3,4\\right)[\/latex]<\/li>\r\n \t<li>[latex]\\left(0,3\\right)[\/latex]<\/li>\r\n \t<li>[latex]\\left(5\\sqrt{3},-5\\right)[\/latex]<\/li>\r\n<\/ol>\r\n<p id=\"fs-id1167793266138\">Convert each of the following points into rectangular coordinates.<\/p>\r\n\r\n<ol id=\"fs-id1167793821570\" start=\"4\" type=\"a\">\r\n \t<li>[latex]\\left(3,\\frac{\\pi}{3}\\right)[\/latex]<\/li>\r\n \t<li>[latex]\\left(2,\\frac{3\\pi}{2}\\right)[\/latex]<\/li>\r\n \t<li>[latex]\\left(6,\\frac{-5\\pi}{6}\\right)[\/latex]<\/li>\r\n<\/ol>\r\n<\/div>\r\n[reveal-answer q=\"44558899\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"44558899\"]\r\n<div id=\"fs-id1167794058406\" data-type=\"solution\">\r\n<ol id=\"fs-id1167794045804\" type=\"a\">\r\n \t<li>Use [latex]x=1[\/latex] and [latex]y=1[\/latex] in the theorem:<span data-type=\"newline\">\r\n<\/span>\r\n<div id=\"fs-id1167793849839\" class=\"unnumbered\" style=\"text-align: center;\" data-type=\"equation\" data-label=\"\">[latex]\\begin{array}{ccccccc}\\begin{array}{ccc}\\hfill {r}^{2}&amp; =\\hfill &amp; {x}^{2}+{y}^{2}\\hfill \\\\ &amp; =\\hfill &amp; {1}^{2}+{1}^{2}\\hfill \\\\ \\hfill r&amp; =\\hfill &amp; \\sqrt{2}\\hfill \\end{array}\\hfill &amp; &amp; &amp; \\text{and}\\hfill &amp; &amp; &amp; \\begin{array}{ccc}\\hfill \\tan\\theta &amp; =\\hfill &amp; \\frac{y}{x}\\hfill \\\\ &amp; =\\hfill &amp; \\frac{1}{1}=1\\hfill \\\\ \\hfill \\theta &amp; =\\hfill &amp; \\frac{\\pi }{4}.\\hfill \\end{array}\\hfill \\end{array}[\/latex]<\/div>\r\n<span data-type=\"newline\">\r\n<\/span>\r\nTherefore this point can be represented as [latex]\\left(\\sqrt{2},\\frac{\\pi }{4}\\right)[\/latex] in polar coordinates.<\/li>\r\n \t<li>Use [latex]x=-3[\/latex] and [latex]y=4[\/latex] in the theorem:<span data-type=\"newline\">\r\n<\/span>\r\n<div id=\"fs-id1167793813480\" class=\"unnumbered\" style=\"text-align: center;\" data-type=\"equation\" data-label=\"\">[latex]\\begin{array}{ccccccc}\\begin{array}{ccc}\\hfill {r}^{2}&amp; =\\hfill &amp; {x}^{2}+{y}^{2}\\hfill \\\\ &amp; =\\hfill &amp; {\\left(-3\\right)}^{2}+{\\left(4\\right)}^{2}\\hfill \\\\ \\hfill r&amp; =\\hfill &amp; 5\\hfill \\end{array}\\hfill &amp; &amp; &amp; \\text{and}\\hfill &amp; &amp; &amp; \\begin{array}{ccc}\\hfill \\tan\\theta &amp; =\\hfill &amp; \\frac{y}{x}\\hfill \\\\ &amp; =\\hfill &amp; -\\frac{4}{3}\\hfill \\\\ \\hfill \\theta &amp; =\\hfill &amp; -\\text{arctan}\\left(\\frac{4}{3}\\right)\\hfill \\\\ &amp; \\approx \\hfill &amp; 2.21.\\hfill \\end{array}\\hfill \\end{array}[\/latex]<\/div>\r\n<span data-type=\"newline\">\r\n<\/span>\r\nTherefore this point can be represented as [latex]\\left(5,2.21\\right)[\/latex] in polar coordinates.<\/li>\r\n \t<li>Use [latex]x=0[\/latex] and [latex]y=3[\/latex] in the theorem:<span data-type=\"newline\">\r\n<\/span>\r\n<div id=\"fs-id1167793819715\" class=\"unnumbered\" style=\"text-align: center;\" data-type=\"equation\" data-label=\"\">[latex]\\begin{array}{ccccccc}\\begin{array}{ccc}\\hfill {r}^{2}&amp; =\\hfill &amp; {x}^{2}+{y}^{2}\\hfill \\\\ &amp; =\\hfill &amp; {\\left(3\\right)}^{2}+{\\left(0\\right)}^{2}\\hfill \\\\ &amp; =\\hfill &amp; 9+0\\hfill \\\\ \\hfill r&amp; =\\hfill &amp; 3\\hfill \\end{array}\\hfill &amp; &amp; &amp; \\text{and}\\hfill &amp; &amp; &amp; \\begin{array}{ccc}\\hfill \\tan\\theta &amp; =\\hfill &amp; \\frac{y}{x}\\hfill \\\\ &amp; =\\hfill &amp; \\frac{3}{0}.\\hfill \\end{array}\\hfill \\end{array}[\/latex]<\/div>\r\n<span data-type=\"newline\">\r\n<\/span>\r\nDirect application of the second equation leads to division by zero. Graphing the point [latex]\\left(0,3\\right)[\/latex] on the rectangular coordinate system reveals that the point is located on the positive [latex]y[\/latex]-axis. The angle between the positive [latex]x[\/latex]-axis and the positive [latex]y[\/latex]-axis is [latex]\\frac{\\pi }{2}[\/latex]. Therefore this point can be represented as [latex]\\left(3,\\frac{\\pi }{2}\\right)[\/latex] in polar coordinates.<\/li>\r\n \t<li>Use [latex]x=5\\sqrt{3}[\/latex] and [latex]y=-5[\/latex] in the theorem:<span data-type=\"newline\">\r\n<\/span>\r\n<div id=\"fs-id1167793878815\" class=\"unnumbered\" style=\"text-align: center;\" data-type=\"equation\" data-label=\"\">[latex]\\begin{array}{ccccccc}\\begin{array}{ccc}\\hfill {r}^{2}&amp; =\\hfill &amp; {x}^{2}+{y}^{2}\\hfill \\\\ &amp; =\\hfill &amp; {\\left(5\\sqrt{3}\\right)}^{2}+{\\left(-5\\right)}^{2}\\hfill \\\\ &amp; =\\hfill &amp; 75+25\\hfill \\\\ \\hfill r&amp; =\\hfill &amp; 10\\hfill \\end{array}\\hfill &amp; &amp; &amp; \\text{and}\\hfill &amp; &amp; &amp; \\begin{array}{ccc}\\hfill \\tan\\theta &amp; =\\hfill &amp; \\frac{y}{x}\\hfill \\\\ &amp; =\\hfill &amp; \\frac{-5}{5\\sqrt{3}}=-\\frac{\\sqrt{3}}{3}\\hfill \\\\ \\hfill \\theta &amp; =\\hfill &amp; -\\frac{\\pi }{6}.\\hfill \\end{array}\\hfill \\end{array}[\/latex]<\/div>\r\n<span data-type=\"newline\">\r\n<\/span>\r\nTherefore this point can be represented as [latex]\\left(10,-\\frac{\\pi }{6}\\right)[\/latex] in polar coordinates.<\/li>\r\n \t<li>Use [latex]r=3[\/latex] and [latex]\\theta =\\frac{\\pi }{3}[\/latex] in the theorem:<span data-type=\"newline\">\r\n<\/span>\r\n<div id=\"fs-id1167794047098\" class=\"unnumbered\" style=\"text-align: center;\" data-type=\"equation\" data-label=\"\">[latex]\\begin{array}{ccccccc}\\begin{array}{ccc}\\hfill x&amp; =\\hfill &amp; r\\cos\\theta \\hfill \\\\ &amp; =\\hfill &amp; 3\\cos\\left(\\frac{\\pi }{3}\\right)\\hfill \\\\ &amp; =\\hfill &amp; 3\\left(\\frac{1}{2}\\right)=\\frac{3}{2}\\hfill \\end{array}\\hfill &amp; &amp; &amp; \\text{and}\\hfill &amp; &amp; &amp; \\begin{array}{ccc}\\hfill y&amp; =\\hfill &amp; r\\sin\\theta \\hfill \\\\ &amp; =\\hfill &amp; 3\\sin\\left(\\frac{\\pi }{3}\\right)\\hfill \\\\ &amp; =\\hfill &amp; 3\\left(\\frac{\\sqrt{3}}{2}\\right)=\\frac{3\\sqrt{3}}{2}.\\hfill \\end{array}\\hfill \\end{array}[\/latex]<\/div>\r\n<span data-type=\"newline\">\r\n<\/span>\r\nTherefore this point can be represented as [latex]\\left(\\frac{3}{2},\\frac{3\\sqrt{3}}{2}\\right)[\/latex] in rectangular coordinates.<\/li>\r\n \t<li>Use [latex]r=2[\/latex] and [latex]\\theta =\\frac{3\\pi }{2}[\/latex] in the theorem:<span data-type=\"newline\">\r\n<\/span>\r\n<div id=\"fs-id1167794336338\" class=\"unnumbered\" style=\"text-align: center;\" data-type=\"equation\" data-label=\"\">[latex]\\begin{array}{ccccccc}\\begin{array}{ccc}\\hfill x&amp; =\\hfill &amp; r\\cos\\theta \\hfill \\\\ &amp; =\\hfill &amp; 2\\cos\\left(\\frac{3\\pi }{2}\\right)\\hfill \\\\ &amp; =\\hfill &amp; 2\\left(0\\right)=0\\hfill \\end{array}\\hfill &amp; &amp; &amp; \\text{and}\\hfill &amp; &amp; &amp; \\begin{array}{ccc}\\hfill y&amp; =\\hfill &amp; r\\sin\\theta \\hfill \\\\ &amp; =\\hfill &amp; 2\\sin\\left(\\frac{3\\pi }{2}\\right)\\hfill \\\\ &amp; =\\hfill &amp; 2\\left(-1\\right)=-2.\\hfill \\end{array}\\hfill \\end{array}[\/latex]<\/div>\r\n<span data-type=\"newline\">\r\n<\/span>\r\nTherefore this point can be represented as [latex]\\left(0,-2\\right)[\/latex] in rectangular coordinates.<\/li>\r\n \t<li>Use [latex]r=6[\/latex] and [latex]\\theta =-\\frac{5\\pi }{6}[\/latex] in the theorem:<span data-type=\"newline\">\r\n<\/span>\r\n<div id=\"fs-id1167794047981\" class=\"unnumbered\" style=\"text-align: center;\" data-type=\"equation\" data-label=\"\">[latex]\\begin{array}{ccccccc}\\begin{array}{ccc}\\hfill x&amp; =\\hfill &amp; r\\cos\\theta \\hfill \\\\ &amp; =\\hfill &amp; 6\\cos\\left(-\\frac{5\\pi }{6}\\right)\\hfill \\\\ &amp; =\\hfill &amp; 6\\left(-\\frac{\\sqrt{3}}{2}\\right)\\hfill \\\\ &amp; =\\hfill &amp; -3\\sqrt{3}\\hfill \\end{array}\\hfill &amp; &amp; &amp; \\text{and}\\hfill &amp; &amp; &amp; \\begin{array}{ccc}\\hfill y&amp; =\\hfill &amp; r\\sin\\theta \\hfill \\\\ &amp; =\\hfill &amp; 6\\sin\\left(-\\frac{5\\pi }{6}\\right)\\hfill \\\\ &amp; =\\hfill &amp; 6\\left(-\\frac{1}{2}\\right)\\hfill \\\\ &amp; =\\hfill &amp; -3.\\hfill \\end{array}\\hfill \\end{array}[\/latex]<\/div>\r\n<span data-type=\"newline\">\r\n<\/span>\r\nTherefore this point can be represented as [latex]\\left(-3\\sqrt{3},-3\\right)[\/latex] in rectangular coordinates.<\/li>\r\n<\/ol>\r\n<\/div>\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<\/div>\r\n<\/div>\r\n<\/div>\r\n<\/div>\r\nWatch the following video to see the worked solution to Example: Converting between Rectangular and Polar Coordinates.\r\n\r\n<center><iframe title=\"YouTube video player\" src=\"https:\/\/www.youtube.com\/embed\/m_wIuLZn03U?controls=0&amp;start=72&amp;end=510&amp;autoplay=0\" width=\"750\" height=\"450\" frameborder=\"0\" allowfullscreen=\"allowfullscreen\"><\/iframe><\/center>\r\n<p class=\"p1\">For closed captioning, open the video on its original page by clicking the Youtube logo in the lower right-hand corner of the video display. In YouTube, the video will begin at the same starting point as this clip, but will continue playing until the very end.<\/p>\r\nYou can view the <a href=\"https:\/\/oerfiles.s3.us-west-2.amazonaws.com\/Calculus+II\/Transcripts\/7.3PolarCoordinates72to510_transcript.html\" target=\"_blank\" rel=\"noopener\">transcript for this segmented clip of \"7.3 Polar Coordinates\" here (opens in new window)<\/a>.\r\n<div id=\"fs-id1167793901804\" class=\"checkpoint\" data-type=\"note\">\r\n<div id=\"fs-id1167793901807\" data-type=\"exercise\">\r\n<div id=\"fs-id1167793901809\" data-type=\"problem\">\r\n<div class=\"textbox key-takeaways\">\r\n<h3>try it<\/h3>\r\n<div id=\"fs-id1167793901809\" data-type=\"problem\">\r\n<p id=\"fs-id1167794066199\">Convert [latex]\\left(-8,-8\\right)[\/latex] into polar coordinates and [latex]\\left(4,\\frac{2\\pi }{3}\\right)[\/latex] into rectangular coordinates.<\/p>\r\n\r\n<\/div>\r\n[reveal-answer q=\"44558897\"]Hint[\/reveal-answer]\r\n[hidden-answer a=\"44558897\"]\r\n<div id=\"fs-id1167794064210\" data-type=\"commentary\" data-element-type=\"hint\">\r\n<p id=\"fs-id1167793262229\">Use both equations from the theorem. Make sure to check the quadrant when calculating [latex]\\theta [\/latex].<\/p>\r\n\r\n<\/div>\r\n[\/hidden-answer]\r\n\r\n[reveal-answer q=\"44558898\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"44558898\"]\r\n<div id=\"fs-id1167793805544\" data-type=\"solution\">\r\n<p id=\"fs-id1167793270874\" style=\"text-align: center;\">[latex]\\left(8\\sqrt{2},\\frac{5\\pi }{4}\\right)[\/latex] and [latex]\\left(-2,2\\sqrt{3}\\right)[\/latex]<\/p>\r\n\r\n<\/div>\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<\/div>\r\n<\/div>\r\n<\/div>\r\n<div class=\"textbox key-takeaways\">\r\n<h3>Try It<\/h3>\r\n[ohm_question]8488[\/ohm_question]\r\n\r\n<\/div>\r\n<p id=\"fs-id1167794048044\">The polar representation of a point is not unique. For example, the polar coordinates [latex]\\left(2,\\frac{\\pi }{3}\\right)[\/latex] and [latex]\\left(2,\\frac{7\\pi }{3}\\right)[\/latex] both represent the point [latex]\\left(1,\\sqrt{3}\\right)[\/latex] in the rectangular system. Also, the value of [latex]r[\/latex] can be negative. Therefore, the point with polar coordinates [latex]\\left(-2,\\frac{4\\pi }{3}\\right)[\/latex] also represents the point [latex]\\left(1,\\sqrt{3}\\right)[\/latex] in the rectangular system, as we can see by using the theorem:<\/p>\r\n\r\n<div id=\"fs-id1167793261943\" class=\"unnumbered\" style=\"text-align: center;\" data-type=\"equation\" data-label=\"\">[latex]\\begin{array}{ccccccc}\\begin{array}{ccc}\\hfill x&amp; =\\hfill &amp; r\\cos\\theta \\hfill \\\\ &amp; =\\hfill &amp; -2\\cos\\left(\\frac{4\\pi }{3}\\right)\\hfill \\\\ &amp; =\\hfill &amp; -2\\left(-\\frac{1}{2}\\right)=1\\hfill \\end{array}\\hfill &amp; &amp; &amp; \\text{and}\\hfill &amp; &amp; &amp; \\begin{array}{ccc}\\hfill y&amp; =\\hfill &amp; r\\sin\\theta \\hfill \\\\ &amp; =\\hfill &amp; -2\\sin\\left(\\frac{4\\pi }{3}\\right)\\hfill \\\\ &amp; =\\hfill &amp; -2\\left(-\\frac{\\sqrt{3}}{2}\\right)=\\sqrt{3}.\\hfill \\end{array}\\hfill \\end{array}[\/latex]<\/div>\r\n&nbsp;\r\n<p id=\"fs-id1167793269541\">Every point in the plane has an infinite number of representations in polar coordinates. However, each point in the plane has only one representation in the rectangular coordinate system.<\/p>\r\n<p id=\"fs-id1167793269546\">Note that the polar representation of a point in the plane also has a visual interpretation. In particular, [latex]r[\/latex] is the directed distance that the point lies from the origin, and [latex]\\theta [\/latex] measures the angle that the line segment from the origin to the point makes with the positive [latex]x[\/latex] -axis. Positive angles are measured in a counterclockwise direction and negative angles are measured in a clockwise direction. The polar coordinate system appears in the following figure.<\/p>\r\n\r\n<figure id=\"CNX_Calc_Figure_11_03_002\"><figcaption><\/figcaption>[caption id=\"\" align=\"aligncenter\" width=\"576\"]<img src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/4175\/2019\/04\/11234809\/CNX_Calc_Figure_11_03_002.jpg\" alt=\"A series of concentric circles is drawn with spokes indicating different values between 0 and 2\u03c0 in increments of \u03c0\/12. The first quadrant starts with 0 where the [latex]x[\/latex]-axis would be, then the next spoke is marked \u03c0\/12, then \u03c0\/6, \u03c0\/4, \u03c0\/3, 5\u03c0\/12, \u03c0\/2, and so on into the second, third, and fourth quadrants. The polar axis is noted near the former [latex]x[\/latex]-axis line.\" width=\"576\" height=\"499\" data-media-type=\"image\/jpeg\" \/> Figure 2. The polar coordinate system.[\/caption]<\/figure>\r\n<p id=\"fs-id1167794210902\">The line segment starting from the center of the graph going to the right (called the positive [latex]x[\/latex]-axis in the Cartesian system) is the <span data-type=\"term\">polar axis<\/span>. The center point is the <span data-type=\"term\">pole<\/span>, or origin, of the coordinate system, and corresponds to [latex]r=0[\/latex]. The innermost circle shown in Figure 2 contains all points a distance of 1 unit from the pole, and is represented by the equation [latex]r=1[\/latex]. Then [latex]r=2[\/latex] is the set of points 2 units from the pole, and so on. The line segments emanating from the pole correspond to fixed angles. To plot a point in the polar coordinate system, start with the angle. If the angle is positive, then measure the angle from the polar axis in a counterclockwise direction. If it is negative, then measure it clockwise. If the value of [latex]r[\/latex] is positive, move that distance along the terminal ray of the angle. If it is negative, move along the ray that is opposite the terminal ray of the given angle.<\/p>\r\n\r\n<div id=\"fs-id1167794066465\" data-type=\"example\">\r\n<div id=\"fs-id1167794066467\" data-type=\"exercise\">\r\n<div id=\"fs-id1167793251138\" data-type=\"problem\">\r\n<div data-type=\"title\">\r\n<div class=\"textbox exercises\">\r\n<h3>Example: Plotting Points in the Polar Plane<\/h3>\r\n<div id=\"fs-id1167793251138\" data-type=\"problem\">\r\n<p id=\"fs-id1167793251143\">Plot each of the following points on the polar plane.<\/p>\r\n\r\n<ol id=\"fs-id1167793251146\" type=\"a\">\r\n \t<li>[latex]\\left(2,\\frac{\\pi }{4}\\right)[\/latex]<\/li>\r\n \t<li>[latex]\\left(-3,\\frac{2\\pi }{3}\\right)[\/latex]<\/li>\r\n \t<li>[latex]\\left(4,\\frac{5\\pi }{4}\\right)[\/latex]<\/li>\r\n<\/ol>\r\n<\/div>\r\n[reveal-answer q=\"44558896\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"44558896\"]\r\n<div id=\"fs-id1167794094091\" data-type=\"solution\">\r\n<p id=\"fs-id1167794332429\">The three points are plotted in the following figure.<\/p>\r\n\r\n<figure id=\"CNX_Calc_Figure_11_03_003\"><figcaption><\/figcaption>\r\n\r\n[caption id=\"\" align=\"aligncenter\" width=\"417\"]<img src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/4175\/2019\/04\/11234812\/CNX_Calc_Figure_11_03_003.jpg\" alt=\"Three points are marked on a polar coordinate plane, specifically (2, \u03c0\/4) in the first quadrant, (4, 5\u03c0\/4) in the third quadrant, and (\u22123, 2\u03c0\/3) in the fourth quadrant.\" width=\"417\" height=\"417\" data-media-type=\"image\/jpeg\" \/> Figure 3. Three points plotted in the polar coordinate system.[\/caption]<\/figure>\r\n<\/div>\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<\/div>\r\n<\/div>\r\n<\/div>\r\n<\/div>\r\nWatch the following video to see the worked solution to Example: Plotting Points in the Polar Plane.\r\n\r\n<center><iframe title=\"YouTube video player\" src=\"https:\/\/www.youtube.com\/embed\/m_wIuLZn03U?controls=0&amp;start=557&amp;end=664&amp;autoplay=0\" width=\"750\" height=\"450\" frameborder=\"0\" allowfullscreen=\"allowfullscreen\"><\/iframe><\/center>\r\n<p class=\"p1\">For closed captioning, open the video on its original page by clicking the Youtube logo in the lower right-hand corner of the video display. In YouTube, the video will begin at the same starting point as this clip, but will continue playing until the very end.<\/p>\r\nYou can view the <a href=\"https:\/\/oerfiles.s3.us-west-2.amazonaws.com\/Calculus+II\/Transcripts\/7.3PolarCoordinates557to664_transcript.html\" target=\"_blank\" rel=\"noopener\">transcript for this segmented clip of \"7.3 Polar Coordinates\" here (opens in new window)<\/a>.\r\n<div id=\"fs-id1167794292518\" class=\"checkpoint\" data-type=\"note\">\r\n<div id=\"fs-id1167794292522\" data-type=\"exercise\">\r\n<div id=\"fs-id1167794292524\" data-type=\"problem\">\r\n<div class=\"textbox key-takeaways\">\r\n<h3>try it<\/h3>\r\n<div id=\"fs-id1167794292524\" data-type=\"problem\">\r\n<p id=\"fs-id1167794330185\">Plot [latex]\\left(4,\\frac{5\\pi }{3}\\right)[\/latex] and [latex]\\left(-3,-\\frac{7\\pi }{2}\\right)[\/latex] on the polar plane.<\/p>\r\n\r\n<\/div>\r\n[reveal-answer q=\"44558894\"]Hint[\/reveal-answer]\r\n[hidden-answer a=\"44558894\"]\r\n<div id=\"fs-id1167794039158\" data-type=\"commentary\" data-element-type=\"hint\">\r\n<p id=\"fs-id1167794060098\">Start with [latex]\\theta [\/latex], then use [latex]r[\/latex].<\/p>\r\n\r\n<\/div>\r\n[\/hidden-answer]\r\n\r\n[reveal-answer q=\"44558895\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"44558895\"]\r\n<div id=\"fs-id1167794039142\" data-type=\"solution\">\r\n<p id=\"fs-id1167794039143\"><span data-type=\"newline\">\u00a0<\/span><\/p>\r\n\r\n[caption id=\"\" align=\"aligncenter\" width=\"417\"]<img src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/4175\/2019\/04\/11234815\/CNX_Calc_Figure_11_03_004.jpg\" alt=\"Two points are marked on a polar coordinate plane, specifically (\u22123, \u22127\u03c0\/2) on the [latex]y[\/latex]-axis and (4, 5\u03c0\/3) in the fourth quadrant.\" width=\"417\" height=\"417\" data-media-type=\"image\/jpeg\" \/> Figure 4.[\/caption]<\/div>\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<\/div>\r\n<\/div>\r\n<\/div>\r\n<\/section><section id=\"fs-id1167793830753\" data-depth=\"1\"><\/section>","rendered":"<div class=\"textbox learning-objectives\" data-type=\"abstract\">\n<h3>Learning Outcomes<\/h3>\n<ul>\n<li>Locate points in a plane by using polar coordinates<\/li>\n<li>Convert points between rectangular and polar coordinates<\/li>\n<\/ul>\n<\/div>\n<section id=\"fs-id1167794030048\" data-depth=\"1\">\n<p id=\"fs-id1167794039551\">To find the coordinates of a point in the polar coordinate system, consider Figure 1. The point [latex]P[\/latex] has Cartesian coordinates [latex]\\left(x,y\\right)[\/latex]. The line segment connecting the origin to the point [latex]P[\/latex] measures the distance from the origin to [latex]P[\/latex] and has length [latex]r[\/latex]. The angle between the positive [latex]x[\/latex] -axis and the line segment has measure [latex]\\theta[\/latex]. This observation suggests a natural correspondence between the coordinate pair [latex]\\left(x,y\\right)[\/latex] and the values [latex]r[\/latex] and [latex]\\theta[\/latex]. This correspondence is the basis of the <strong>polar coordinate system<\/strong>. Note that every point in the Cartesian plane has two values (hence the term <em data-effect=\"italics\">ordered pair<\/em>) associated with it. In the polar coordinate system, each point also has two values associated with it: [latex]r[\/latex] and [latex]\\theta[\/latex].<\/p>\n<figure id=\"CNX_Calc_Figure_11_03_001\"><figcaption><\/figcaption><div style=\"width: 314px\" class=\"wp-caption aligncenter\"><img loading=\"lazy\" decoding=\"async\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/4175\/2019\/04\/11234807\/CNX_Calc_Figure_11_03_001.jpg\" alt=\"A point P(x, y) is given in the first quadrant with lines drawn to indicate its x and y values. There is a line from the origin to P(x, y) marked r and this line make an angle \u03b8 with the x axis.\" width=\"304\" height=\"309\" data-media-type=\"image\/jpeg\" \/><\/p>\n<p class=\"wp-caption-text\">Figure 1. An arbitrary point in the Cartesian plane.<\/p>\n<\/div>\n<\/figure>\n<p>As we can observe above, the right triangle pictured above implies a set of relationships between [latex]x, y, r, \\:\\text{and}\\: \\theta[\/latex].\u00a0 We first recall those relationships below.<\/p>\n<div class=\"textbox examples\">\n<h3>Recall: Right Triangle Trigonometry<\/h3>\n<p>Given a right triangle with an acute angle of [latex]\\theta[\/latex],<\/p>\n<p style=\"text-align: left; padding-left: 60px;\">[latex]\\begin{align}&\\sin \\left(\\theta \\right)=\\frac{\\text{opposite}}{\\text{hypotenuse}} \\\\ &\\cos \\left(\\theta \\right)=\\frac{\\text{adjacent}}{\\text{hypotenuse}} \\\\ &\\tan \\left(\\theta \\right)=\\frac{\\text{opposite}}{\\text{adjacent}} \\end{align}[\/latex]<\/p>\n<p>A common mnemonic for remembering these relationships is SohCahToa, formed from the first letters of &#8220;Sine is opposite over hypotenuse, Cosine is adjacent over hypotenuse, Tangent is opposite over adjacent.&#8221;<\/p>\n<p>The side lengths of the right triangle with legs [latex]a[\/latex] and [latex]b[\/latex] and hypotenuse\u00a0[latex]c[\/latex] are related through the Pythagorean Theorem:\u00a0[latex]a^2 + b^2 = c^2[\/latex]<\/p>\n<\/div>\n<p id=\"fs-id1167794070618\">Using right-triangle trigonometry, the following equations are true for the point [latex]P\\text{:}[\/latex]<\/p>\n<div id=\"fs-id1167794187190\" class=\"unnumbered\" style=\"text-align: center;\" data-type=\"equation\" data-label=\"\">[latex]\\cos\\theta =\\frac{x}{r}\\:\\:\\text{so}\\:\\:x=r\\cos\\theta[\/latex]<\/div>\n<p>&nbsp;<\/p>\n<div id=\"fs-id1167794170718\" class=\"unnumbered\" style=\"text-align: center;\" data-type=\"equation\" data-label=\"\">[latex]\\sin\\theta =\\frac{y}{r}\\:\\:\\text{ so }\\:\\:y=r\\sin\\theta[\/latex].<\/div>\n<p>&nbsp;<\/p>\n<p id=\"fs-id1167793785830\">Furthermore,<\/p>\n<div id=\"fs-id1167793821731\" class=\"unnumbered\" style=\"text-align: center;\" data-type=\"equation\" data-label=\"\">[latex]{r}^{2}={x}^{2}+{y}^{2}\\text{ and }\\tan\\theta =\\frac{y}{x}[\/latex].<\/div>\n<p>&nbsp;<\/p>\n<p id=\"fs-id1167794056124\">Each point [latex]\\left(x,y\\right)[\/latex] in the Cartesian coordinate system can therefore be represented as an ordered pair [latex]\\left(r,\\theta \\right)[\/latex] in the polar coordinate system. The first coordinate is called the <span data-type=\"term\">radial coordinate<\/span> and the second coordinate is called the <span data-type=\"term\">angular coordinate<\/span>. Every point in the plane can be represented in this form.<\/p>\n<p id=\"fs-id1167794170727\">Note that the equation [latex]\\tan\\theta =\\frac{y}{x}[\/latex] has an infinite number of solutions for any ordered pair [latex]\\left(x,y\\right)[\/latex]. However, if we restrict the solutions to values between [latex]0[\/latex] and [latex]2\\pi[\/latex] then we can assign a unique solution to the quadrant in which the original point [latex]\\left(x,y\\right)[\/latex] is located. Then the corresponding value of <em data-effect=\"italics\">r<\/em> is positive, so [latex]{r}^{2}={x}^{2}+{y}^{2}[\/latex].<\/p>\n<div id=\"fs-id1167793948751\" class=\"theorem\" data-type=\"note\">\n<div data-type=\"title\">\n<div class=\"textbox shaded\">\n<h3 style=\"text-align: center;\" data-type=\"title\">theorem: Converting Points between Coordinate Systems<\/h3>\n<hr \/>\n<p id=\"fs-id1167794024884\">Given a point [latex]P[\/latex] in the plane with Cartesian coordinates [latex]\\left(x,y\\right)[\/latex] and polar coordinates [latex]\\left(r,\\theta \\right)[\/latex], the following conversion formulas hold true:<\/p>\n<div id=\"fs-id1167794066096\" style=\"text-align: center;\" data-type=\"equation\">[latex]x=r\\cos\\theta \\:\\text{and}\\:y=r\\sin\\theta[\/latex],<\/div>\n<p>&nbsp;<\/p>\n<div id=\"fs-id1167793834620\" style=\"text-align: center;\" data-type=\"equation\">[latex]{r}^{2}={x}^{2}+{y}^{2}\\:\\text{and}\\:\\tan\\theta =\\frac{y}{x}[\/latex].<\/div>\n<p>&nbsp;<\/p>\n<p id=\"fs-id1167793913264\">These formulas can be used to convert from rectangular to polar or from polar to rectangular coordinates.<\/p>\n<\/div>\n<\/div>\n<\/div>\n<p>As we can note carefully above, the equation [latex]\\tan\\theta =\\frac{y}{x}[\/latex] is <em>not<\/em> expressed using the inverse tangent function.\u00a0 The reason for this is because of how the domain restriction of the tangent function leads to a restricted range of the inverse tangent function, reviewed below.<\/p>\n<div class=\"textbox examples\">\n<h3>Recall: Using The Inverse Tangent Function in the coordinate plane<\/h3>\n<p>If [latex]-\\frac{\\pi}{2} < \\theta < \\frac{\\pi}{2}[\/latex], and [latex]\\tan \\theta = \\frac{y}{x}[\/latex], then [latex]\\theta = \\tan^{-1} \\left( \\frac{y}{x} \\right)[\/latex]\n\n\n<p style=\"padding-left: 30px;\">That is, the inverse tangent function has a range of [latex]\\left(-\\frac{\\pi}{2}, \\frac{\\pi}{2}\\right)[\/latex], meaning that it always produces a positive angle in Quadrant I or a negative angle in Quadrant IV.<\/p>\n<p>If [latex]\\frac{\\pi}{2} < \\theta < \\frac{3\\pi}{2}[\/latex] and [latex]\\tan \\theta = \\frac{y}{x}[\/latex], then [latex]\\theta = \\tan^{-1} \\left( \\frac{y}{x} \\right) + \\pi[\/latex]\n\n\n<p style=\"padding-left: 30px;\">In other words, if the point [latex]\\left(x, y \\right)[\/latex] is in Quadrant II or III, the preceding rule means that you must add [latex]\\pi[\/latex] to the output of the inverse tangent function to produce an angle in the correct quadrant.<\/p>\n<\/div>\n<div id=\"fs-id1167794030722\" data-type=\"example\">\n<div id=\"fs-id1167794003809\" data-type=\"exercise\">\n<div id=\"fs-id1167794044637\" data-type=\"problem\">\n<div data-type=\"title\">\n<div class=\"textbox exercises\">\n<h3>Example: Converting between Rectangular and Polar Coordinates<\/h3>\n<div id=\"fs-id1167794044637\" data-type=\"problem\">\n<p id=\"fs-id1167793814959\">Convert each of the following points into polar coordinates.<\/p>\n<ol id=\"fs-id1167793906173\" type=\"a\">\n<li>[latex]\\left(1,1\\right)[\/latex]<\/li>\n<li>[latex]\\left(-3,4\\right)[\/latex]<\/li>\n<li>[latex]\\left(0,3\\right)[\/latex]<\/li>\n<li>[latex]\\left(5\\sqrt{3},-5\\right)[\/latex]<\/li>\n<\/ol>\n<p id=\"fs-id1167793266138\">Convert each of the following points into rectangular coordinates.<\/p>\n<ol id=\"fs-id1167793821570\" start=\"4\" type=\"a\">\n<li>[latex]\\left(3,\\frac{\\pi}{3}\\right)[\/latex]<\/li>\n<li>[latex]\\left(2,\\frac{3\\pi}{2}\\right)[\/latex]<\/li>\n<li>[latex]\\left(6,\\frac{-5\\pi}{6}\\right)[\/latex]<\/li>\n<\/ol>\n<\/div>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q44558899\">Show Solution<\/span><\/p>\n<div id=\"q44558899\" class=\"hidden-answer\" style=\"display: none\">\n<div id=\"fs-id1167794058406\" data-type=\"solution\">\n<ol id=\"fs-id1167794045804\" type=\"a\">\n<li>Use [latex]x=1[\/latex] and [latex]y=1[\/latex] in the theorem:<span data-type=\"newline\"><br \/>\n<\/span><\/p>\n<div id=\"fs-id1167793849839\" class=\"unnumbered\" style=\"text-align: center;\" data-type=\"equation\" data-label=\"\">[latex]\\begin{array}{ccccccc}\\begin{array}{ccc}\\hfill {r}^{2}& =\\hfill & {x}^{2}+{y}^{2}\\hfill \\\\ & =\\hfill & {1}^{2}+{1}^{2}\\hfill \\\\ \\hfill r& =\\hfill & \\sqrt{2}\\hfill \\end{array}\\hfill & & & \\text{and}\\hfill & & & \\begin{array}{ccc}\\hfill \\tan\\theta & =\\hfill & \\frac{y}{x}\\hfill \\\\ & =\\hfill & \\frac{1}{1}=1\\hfill \\\\ \\hfill \\theta & =\\hfill & \\frac{\\pi }{4}.\\hfill \\end{array}\\hfill \\end{array}[\/latex]<\/div>\n<p><span data-type=\"newline\"><br \/>\n<\/span><br \/>\nTherefore this point can be represented as [latex]\\left(\\sqrt{2},\\frac{\\pi }{4}\\right)[\/latex] in polar coordinates.<\/li>\n<li>Use [latex]x=-3[\/latex] and [latex]y=4[\/latex] in the theorem:<span data-type=\"newline\"><br \/>\n<\/span><\/p>\n<div id=\"fs-id1167793813480\" class=\"unnumbered\" style=\"text-align: center;\" data-type=\"equation\" data-label=\"\">[latex]\\begin{array}{ccccccc}\\begin{array}{ccc}\\hfill {r}^{2}& =\\hfill & {x}^{2}+{y}^{2}\\hfill \\\\ & =\\hfill & {\\left(-3\\right)}^{2}+{\\left(4\\right)}^{2}\\hfill \\\\ \\hfill r& =\\hfill & 5\\hfill \\end{array}\\hfill & & & \\text{and}\\hfill & & & \\begin{array}{ccc}\\hfill \\tan\\theta & =\\hfill & \\frac{y}{x}\\hfill \\\\ & =\\hfill & -\\frac{4}{3}\\hfill \\\\ \\hfill \\theta & =\\hfill & -\\text{arctan}\\left(\\frac{4}{3}\\right)\\hfill \\\\ & \\approx \\hfill & 2.21.\\hfill \\end{array}\\hfill \\end{array}[\/latex]<\/div>\n<p><span data-type=\"newline\"><br \/>\n<\/span><br \/>\nTherefore this point can be represented as [latex]\\left(5,2.21\\right)[\/latex] in polar coordinates.<\/li>\n<li>Use [latex]x=0[\/latex] and [latex]y=3[\/latex] in the theorem:<span data-type=\"newline\"><br \/>\n<\/span><\/p>\n<div id=\"fs-id1167793819715\" class=\"unnumbered\" style=\"text-align: center;\" data-type=\"equation\" data-label=\"\">[latex]\\begin{array}{ccccccc}\\begin{array}{ccc}\\hfill {r}^{2}& =\\hfill & {x}^{2}+{y}^{2}\\hfill \\\\ & =\\hfill & {\\left(3\\right)}^{2}+{\\left(0\\right)}^{2}\\hfill \\\\ & =\\hfill & 9+0\\hfill \\\\ \\hfill r& =\\hfill & 3\\hfill \\end{array}\\hfill & & & \\text{and}\\hfill & & & \\begin{array}{ccc}\\hfill \\tan\\theta & =\\hfill & \\frac{y}{x}\\hfill \\\\ & =\\hfill & \\frac{3}{0}.\\hfill \\end{array}\\hfill \\end{array}[\/latex]<\/div>\n<p><span data-type=\"newline\"><br \/>\n<\/span><br \/>\nDirect application of the second equation leads to division by zero. Graphing the point [latex]\\left(0,3\\right)[\/latex] on the rectangular coordinate system reveals that the point is located on the positive [latex]y[\/latex]-axis. The angle between the positive [latex]x[\/latex]-axis and the positive [latex]y[\/latex]-axis is [latex]\\frac{\\pi }{2}[\/latex]. Therefore this point can be represented as [latex]\\left(3,\\frac{\\pi }{2}\\right)[\/latex] in polar coordinates.<\/li>\n<li>Use [latex]x=5\\sqrt{3}[\/latex] and [latex]y=-5[\/latex] in the theorem:<span data-type=\"newline\"><br \/>\n<\/span><\/p>\n<div id=\"fs-id1167793878815\" class=\"unnumbered\" style=\"text-align: center;\" data-type=\"equation\" data-label=\"\">[latex]\\begin{array}{ccccccc}\\begin{array}{ccc}\\hfill {r}^{2}& =\\hfill & {x}^{2}+{y}^{2}\\hfill \\\\ & =\\hfill & {\\left(5\\sqrt{3}\\right)}^{2}+{\\left(-5\\right)}^{2}\\hfill \\\\ & =\\hfill & 75+25\\hfill \\\\ \\hfill r& =\\hfill & 10\\hfill \\end{array}\\hfill & & & \\text{and}\\hfill & & & \\begin{array}{ccc}\\hfill \\tan\\theta & =\\hfill & \\frac{y}{x}\\hfill \\\\ & =\\hfill & \\frac{-5}{5\\sqrt{3}}=-\\frac{\\sqrt{3}}{3}\\hfill \\\\ \\hfill \\theta & =\\hfill & -\\frac{\\pi }{6}.\\hfill \\end{array}\\hfill \\end{array}[\/latex]<\/div>\n<p><span data-type=\"newline\"><br \/>\n<\/span><br \/>\nTherefore this point can be represented as [latex]\\left(10,-\\frac{\\pi }{6}\\right)[\/latex] in polar coordinates.<\/li>\n<li>Use [latex]r=3[\/latex] and [latex]\\theta =\\frac{\\pi }{3}[\/latex] in the theorem:<span data-type=\"newline\"><br \/>\n<\/span><\/p>\n<div id=\"fs-id1167794047098\" class=\"unnumbered\" style=\"text-align: center;\" data-type=\"equation\" data-label=\"\">[latex]\\begin{array}{ccccccc}\\begin{array}{ccc}\\hfill x& =\\hfill & r\\cos\\theta \\hfill \\\\ & =\\hfill & 3\\cos\\left(\\frac{\\pi }{3}\\right)\\hfill \\\\ & =\\hfill & 3\\left(\\frac{1}{2}\\right)=\\frac{3}{2}\\hfill \\end{array}\\hfill & & & \\text{and}\\hfill & & & \\begin{array}{ccc}\\hfill y& =\\hfill & r\\sin\\theta \\hfill \\\\ & =\\hfill & 3\\sin\\left(\\frac{\\pi }{3}\\right)\\hfill \\\\ & =\\hfill & 3\\left(\\frac{\\sqrt{3}}{2}\\right)=\\frac{3\\sqrt{3}}{2}.\\hfill \\end{array}\\hfill \\end{array}[\/latex]<\/div>\n<p><span data-type=\"newline\"><br \/>\n<\/span><br \/>\nTherefore this point can be represented as [latex]\\left(\\frac{3}{2},\\frac{3\\sqrt{3}}{2}\\right)[\/latex] in rectangular coordinates.<\/li>\n<li>Use [latex]r=2[\/latex] and [latex]\\theta =\\frac{3\\pi }{2}[\/latex] in the theorem:<span data-type=\"newline\"><br \/>\n<\/span><\/p>\n<div id=\"fs-id1167794336338\" class=\"unnumbered\" style=\"text-align: center;\" data-type=\"equation\" data-label=\"\">[latex]\\begin{array}{ccccccc}\\begin{array}{ccc}\\hfill x& =\\hfill & r\\cos\\theta \\hfill \\\\ & =\\hfill & 2\\cos\\left(\\frac{3\\pi }{2}\\right)\\hfill \\\\ & =\\hfill & 2\\left(0\\right)=0\\hfill \\end{array}\\hfill & & & \\text{and}\\hfill & & & \\begin{array}{ccc}\\hfill y& =\\hfill & r\\sin\\theta \\hfill \\\\ & =\\hfill & 2\\sin\\left(\\frac{3\\pi }{2}\\right)\\hfill \\\\ & =\\hfill & 2\\left(-1\\right)=-2.\\hfill \\end{array}\\hfill \\end{array}[\/latex]<\/div>\n<p><span data-type=\"newline\"><br \/>\n<\/span><br \/>\nTherefore this point can be represented as [latex]\\left(0,-2\\right)[\/latex] in rectangular coordinates.<\/li>\n<li>Use [latex]r=6[\/latex] and [latex]\\theta =-\\frac{5\\pi }{6}[\/latex] in the theorem:<span data-type=\"newline\"><br \/>\n<\/span><\/p>\n<div id=\"fs-id1167794047981\" class=\"unnumbered\" style=\"text-align: center;\" data-type=\"equation\" data-label=\"\">[latex]\\begin{array}{ccccccc}\\begin{array}{ccc}\\hfill x& =\\hfill & r\\cos\\theta \\hfill \\\\ & =\\hfill & 6\\cos\\left(-\\frac{5\\pi }{6}\\right)\\hfill \\\\ & =\\hfill & 6\\left(-\\frac{\\sqrt{3}}{2}\\right)\\hfill \\\\ & =\\hfill & -3\\sqrt{3}\\hfill \\end{array}\\hfill & & & \\text{and}\\hfill & & & \\begin{array}{ccc}\\hfill y& =\\hfill & r\\sin\\theta \\hfill \\\\ & =\\hfill & 6\\sin\\left(-\\frac{5\\pi }{6}\\right)\\hfill \\\\ & =\\hfill & 6\\left(-\\frac{1}{2}\\right)\\hfill \\\\ & =\\hfill & -3.\\hfill \\end{array}\\hfill \\end{array}[\/latex]<\/div>\n<p><span data-type=\"newline\"><br \/>\n<\/span><br \/>\nTherefore this point can be represented as [latex]\\left(-3\\sqrt{3},-3\\right)[\/latex] in rectangular coordinates.<\/li>\n<\/ol>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<p>Watch the following video to see the worked solution to Example: Converting between Rectangular and Polar Coordinates.<\/p>\n<div style=\"text-align: center;\"><iframe loading=\"lazy\" title=\"YouTube video player\" src=\"https:\/\/www.youtube.com\/embed\/m_wIuLZn03U?controls=0&amp;start=72&amp;end=510&amp;autoplay=0\" width=\"750\" height=\"450\" frameborder=\"0\" allowfullscreen=\"allowfullscreen\"><\/iframe><\/div>\n<p class=\"p1\">For closed captioning, open the video on its original page by clicking the Youtube logo in the lower right-hand corner of the video display. In YouTube, the video will begin at the same starting point as this clip, but will continue playing until the very end.<\/p>\n<p>You can view the <a href=\"https:\/\/oerfiles.s3.us-west-2.amazonaws.com\/Calculus+II\/Transcripts\/7.3PolarCoordinates72to510_transcript.html\" target=\"_blank\" rel=\"noopener\">transcript for this segmented clip of &#8220;7.3 Polar Coordinates&#8221; here (opens in new window)<\/a>.<\/p>\n<div id=\"fs-id1167793901804\" class=\"checkpoint\" data-type=\"note\">\n<div id=\"fs-id1167793901807\" data-type=\"exercise\">\n<div id=\"fs-id1167793901809\" data-type=\"problem\">\n<div class=\"textbox key-takeaways\">\n<h3>try it<\/h3>\n<div id=\"fs-id1167793901809\" data-type=\"problem\">\n<p id=\"fs-id1167794066199\">Convert [latex]\\left(-8,-8\\right)[\/latex] into polar coordinates and [latex]\\left(4,\\frac{2\\pi }{3}\\right)[\/latex] into rectangular coordinates.<\/p>\n<\/div>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q44558897\">Hint<\/span><\/p>\n<div id=\"q44558897\" class=\"hidden-answer\" style=\"display: none\">\n<div id=\"fs-id1167794064210\" data-type=\"commentary\" data-element-type=\"hint\">\n<p id=\"fs-id1167793262229\">Use both equations from the theorem. Make sure to check the quadrant when calculating [latex]\\theta[\/latex].<\/p>\n<\/div>\n<\/div>\n<\/div>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q44558898\">Show Solution<\/span><\/p>\n<div id=\"q44558898\" class=\"hidden-answer\" style=\"display: none\">\n<div id=\"fs-id1167793805544\" data-type=\"solution\">\n<p id=\"fs-id1167793270874\" style=\"text-align: center;\">[latex]\\left(8\\sqrt{2},\\frac{5\\pi }{4}\\right)[\/latex] and [latex]\\left(-2,2\\sqrt{3}\\right)[\/latex]<\/p>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<div class=\"textbox key-takeaways\">\n<h3>Try It<\/h3>\n<p><iframe loading=\"lazy\" id=\"ohm8488\" class=\"resizable\" src=\"https:\/\/ohm.lumenlearning.com\/multiembedq.php?id=8488&theme=oea&iframe_resize_id=ohm8488&show_question_numbers\" width=\"100%\" height=\"150\"><\/iframe><\/p>\n<\/div>\n<p id=\"fs-id1167794048044\">The polar representation of a point is not unique. For example, the polar coordinates [latex]\\left(2,\\frac{\\pi }{3}\\right)[\/latex] and [latex]\\left(2,\\frac{7\\pi }{3}\\right)[\/latex] both represent the point [latex]\\left(1,\\sqrt{3}\\right)[\/latex] in the rectangular system. Also, the value of [latex]r[\/latex] can be negative. Therefore, the point with polar coordinates [latex]\\left(-2,\\frac{4\\pi }{3}\\right)[\/latex] also represents the point [latex]\\left(1,\\sqrt{3}\\right)[\/latex] in the rectangular system, as we can see by using the theorem:<\/p>\n<div id=\"fs-id1167793261943\" class=\"unnumbered\" style=\"text-align: center;\" data-type=\"equation\" data-label=\"\">[latex]\\begin{array}{ccccccc}\\begin{array}{ccc}\\hfill x& =\\hfill & r\\cos\\theta \\hfill \\\\ & =\\hfill & -2\\cos\\left(\\frac{4\\pi }{3}\\right)\\hfill \\\\ & =\\hfill & -2\\left(-\\frac{1}{2}\\right)=1\\hfill \\end{array}\\hfill & & & \\text{and}\\hfill & & & \\begin{array}{ccc}\\hfill y& =\\hfill & r\\sin\\theta \\hfill \\\\ & =\\hfill & -2\\sin\\left(\\frac{4\\pi }{3}\\right)\\hfill \\\\ & =\\hfill & -2\\left(-\\frac{\\sqrt{3}}{2}\\right)=\\sqrt{3}.\\hfill \\end{array}\\hfill \\end{array}[\/latex]<\/div>\n<p>&nbsp;<\/p>\n<p id=\"fs-id1167793269541\">Every point in the plane has an infinite number of representations in polar coordinates. However, each point in the plane has only one representation in the rectangular coordinate system.<\/p>\n<p id=\"fs-id1167793269546\">Note that the polar representation of a point in the plane also has a visual interpretation. In particular, [latex]r[\/latex] is the directed distance that the point lies from the origin, and [latex]\\theta[\/latex] measures the angle that the line segment from the origin to the point makes with the positive [latex]x[\/latex] -axis. Positive angles are measured in a counterclockwise direction and negative angles are measured in a clockwise direction. The polar coordinate system appears in the following figure.<\/p>\n<figure id=\"CNX_Calc_Figure_11_03_002\"><figcaption><\/figcaption><div style=\"width: 586px\" class=\"wp-caption aligncenter\"><img loading=\"lazy\" decoding=\"async\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/4175\/2019\/04\/11234809\/CNX_Calc_Figure_11_03_002.jpg\" alt=\"A series of concentric circles is drawn with spokes indicating different values between 0 and 2\u03c0 in increments of \u03c0\/12. The first quadrant starts with 0 where the [latex]x[\/latex]-axis would be, then the next spoke is marked \u03c0\/12, then \u03c0\/6, \u03c0\/4, \u03c0\/3, 5\u03c0\/12, \u03c0\/2, and so on into the second, third, and fourth quadrants. The polar axis is noted near the former [latex]x[\/latex]-axis line.\" width=\"576\" height=\"499\" data-media-type=\"image\/jpeg\" \/><\/p>\n<p class=\"wp-caption-text\">Figure 2. The polar coordinate system.<\/p>\n<\/div>\n<\/figure>\n<p id=\"fs-id1167794210902\">The line segment starting from the center of the graph going to the right (called the positive [latex]x[\/latex]-axis in the Cartesian system) is the <span data-type=\"term\">polar axis<\/span>. The center point is the <span data-type=\"term\">pole<\/span>, or origin, of the coordinate system, and corresponds to [latex]r=0[\/latex]. The innermost circle shown in Figure 2 contains all points a distance of 1 unit from the pole, and is represented by the equation [latex]r=1[\/latex]. Then [latex]r=2[\/latex] is the set of points 2 units from the pole, and so on. The line segments emanating from the pole correspond to fixed angles. To plot a point in the polar coordinate system, start with the angle. If the angle is positive, then measure the angle from the polar axis in a counterclockwise direction. If it is negative, then measure it clockwise. If the value of [latex]r[\/latex] is positive, move that distance along the terminal ray of the angle. If it is negative, move along the ray that is opposite the terminal ray of the given angle.<\/p>\n<div id=\"fs-id1167794066465\" data-type=\"example\">\n<div id=\"fs-id1167794066467\" data-type=\"exercise\">\n<div id=\"fs-id1167793251138\" data-type=\"problem\">\n<div data-type=\"title\">\n<div class=\"textbox exercises\">\n<h3>Example: Plotting Points in the Polar Plane<\/h3>\n<div id=\"fs-id1167793251138\" data-type=\"problem\">\n<p id=\"fs-id1167793251143\">Plot each of the following points on the polar plane.<\/p>\n<ol id=\"fs-id1167793251146\" type=\"a\">\n<li>[latex]\\left(2,\\frac{\\pi }{4}\\right)[\/latex]<\/li>\n<li>[latex]\\left(-3,\\frac{2\\pi }{3}\\right)[\/latex]<\/li>\n<li>[latex]\\left(4,\\frac{5\\pi }{4}\\right)[\/latex]<\/li>\n<\/ol>\n<\/div>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q44558896\">Show Solution<\/span><\/p>\n<div id=\"q44558896\" class=\"hidden-answer\" style=\"display: none\">\n<div id=\"fs-id1167794094091\" data-type=\"solution\">\n<p id=\"fs-id1167794332429\">The three points are plotted in the following figure.<\/p>\n<figure id=\"CNX_Calc_Figure_11_03_003\"><figcaption><\/figcaption><div style=\"width: 427px\" class=\"wp-caption aligncenter\"><img loading=\"lazy\" decoding=\"async\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/4175\/2019\/04\/11234812\/CNX_Calc_Figure_11_03_003.jpg\" alt=\"Three points are marked on a polar coordinate plane, specifically (2, \u03c0\/4) in the first quadrant, (4, 5\u03c0\/4) in the third quadrant, and (\u22123, 2\u03c0\/3) in the fourth quadrant.\" width=\"417\" height=\"417\" data-media-type=\"image\/jpeg\" \/><\/p>\n<p class=\"wp-caption-text\">Figure 3. Three points plotted in the polar coordinate system.<\/p>\n<\/div>\n<\/figure>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<p>Watch the following video to see the worked solution to Example: Plotting Points in the Polar Plane.<\/p>\n<div style=\"text-align: center;\"><iframe loading=\"lazy\" title=\"YouTube video player\" src=\"https:\/\/www.youtube.com\/embed\/m_wIuLZn03U?controls=0&amp;start=557&amp;end=664&amp;autoplay=0\" width=\"750\" height=\"450\" frameborder=\"0\" allowfullscreen=\"allowfullscreen\"><\/iframe><\/div>\n<p class=\"p1\">For closed captioning, open the video on its original page by clicking the Youtube logo in the lower right-hand corner of the video display. In YouTube, the video will begin at the same starting point as this clip, but will continue playing until the very end.<\/p>\n<p>You can view the <a href=\"https:\/\/oerfiles.s3.us-west-2.amazonaws.com\/Calculus+II\/Transcripts\/7.3PolarCoordinates557to664_transcript.html\" target=\"_blank\" rel=\"noopener\">transcript for this segmented clip of &#8220;7.3 Polar Coordinates&#8221; here (opens in new window)<\/a>.<\/p>\n<div id=\"fs-id1167794292518\" class=\"checkpoint\" data-type=\"note\">\n<div id=\"fs-id1167794292522\" data-type=\"exercise\">\n<div id=\"fs-id1167794292524\" data-type=\"problem\">\n<div class=\"textbox key-takeaways\">\n<h3>try it<\/h3>\n<div id=\"fs-id1167794292524\" data-type=\"problem\">\n<p id=\"fs-id1167794330185\">Plot [latex]\\left(4,\\frac{5\\pi }{3}\\right)[\/latex] and [latex]\\left(-3,-\\frac{7\\pi }{2}\\right)[\/latex] on the polar plane.<\/p>\n<\/div>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q44558894\">Hint<\/span><\/p>\n<div id=\"q44558894\" class=\"hidden-answer\" style=\"display: none\">\n<div id=\"fs-id1167794039158\" data-type=\"commentary\" data-element-type=\"hint\">\n<p id=\"fs-id1167794060098\">Start with [latex]\\theta[\/latex], then use [latex]r[\/latex].<\/p>\n<\/div>\n<\/div>\n<\/div>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q44558895\">Show Solution<\/span><\/p>\n<div id=\"q44558895\" class=\"hidden-answer\" style=\"display: none\">\n<div id=\"fs-id1167794039142\" data-type=\"solution\">\n<p id=\"fs-id1167794039143\"><span data-type=\"newline\">\u00a0<\/span><\/p>\n<div style=\"width: 427px\" class=\"wp-caption aligncenter\"><img loading=\"lazy\" decoding=\"async\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/4175\/2019\/04\/11234815\/CNX_Calc_Figure_11_03_004.jpg\" alt=\"Two points are marked on a polar coordinate plane, specifically (\u22123, \u22127\u03c0\/2) on the &#091;latex&#093;y&#091;\/latex&#093;-axis and (4, 5\u03c0\/3) in the fourth quadrant.\" width=\"417\" height=\"417\" data-media-type=\"image\/jpeg\" \/><\/p>\n<p class=\"wp-caption-text\">Figure 4.<\/p>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<\/section>\n<section id=\"fs-id1167793830753\" data-depth=\"1\"><\/section>\n\n\t\t\t <section class=\"citations-section\" role=\"contentinfo\">\n\t\t\t <h3>Candela Citations<\/h3>\n\t\t\t\t\t <div>\n\t\t\t\t\t\t <div id=\"citation-list-1161\">\n\t\t\t\t\t\t\t <div class=\"licensing\"><div class=\"license-attribution-dropdown-subheading\">CC licensed content, Original<\/div><ul class=\"citation-list\"><li>7.3 Polar Coordinates. <strong>Authored by<\/strong>: Ryan Melton. <strong>License<\/strong>: <em><a target=\"_blank\" rel=\"license\" href=\"https:\/\/creativecommons.org\/licenses\/by\/4.0\/\">CC BY: Attribution<\/a><\/em><\/li><\/ul><div class=\"license-attribution-dropdown-subheading\">CC licensed content, Shared previously<\/div><ul class=\"citation-list\"><li>Calculus Volume 3. <strong>Authored by<\/strong>: Gilbert Strang, Edwin (Jed) Herman. <strong>Provided by<\/strong>: OpenStax. <strong>Located at<\/strong>: <a target=\"_blank\" href=\"https:\/\/openstax.org\/books\/calculus-volume-3\/pages\/1-introduction\">https:\/\/openstax.org\/books\/calculus-volume-3\/pages\/1-introduction<\/a>. <strong>License<\/strong>: <em><a target=\"_blank\" rel=\"license\" href=\"https:\/\/creativecommons.org\/licenses\/by-nc-sa\/4.0\/\">CC BY-NC-SA: Attribution-NonCommercial-ShareAlike<\/a><\/em>. <strong>License Terms<\/strong>: Access for free at https:\/\/openstax.org\/books\/calculus-volume-3\/pages\/1-introduction<\/li><\/ul><\/div>\n\t\t\t\t\t\t <\/div>\n\t\t\t\t\t <\/div>\n\t\t\t <\/section>","protected":false},"author":349141,"menu_order":11,"template":"","meta":{"_candela_citation":"[{\"type\":\"cc\",\"description\":\"Calculus Volume 3\",\"author\":\"Gilbert Strang, Edwin (Jed) Herman\",\"organization\":\"OpenStax\",\"url\":\"https:\/\/openstax.org\/books\/calculus-volume-3\/pages\/1-introduction\",\"project\":\"\",\"license\":\"cc-by-nc-sa\",\"license_terms\":\"Access for free at https:\/\/openstax.org\/books\/calculus-volume-3\/pages\/1-introduction\"},{\"type\":\"original\",\"description\":\"7.3 Polar Coordinates\",\"author\":\"Ryan Melton\",\"organization\":\"\",\"url\":\"\",\"project\":\"\",\"license\":\"cc-by\",\"license_terms\":\"\"}]","CANDELA_OUTCOMES_GUID":"","pb_show_title":"on","pb_short_title":"","pb_subtitle":"","pb_authors":[],"pb_section_license":""},"chapter-type":[],"contributor":[],"license":[],"class_list":["post-1161","chapter","type-chapter","status-publish","hentry"],"part":1150,"_links":{"self":[{"href":"https:\/\/courses.lumenlearning.com\/calculus3\/wp-json\/pressbooks\/v2\/chapters\/1161","targetHints":{"allow":["GET"]}}],"collection":[{"href":"https:\/\/courses.lumenlearning.com\/calculus3\/wp-json\/pressbooks\/v2\/chapters"}],"about":[{"href":"https:\/\/courses.lumenlearning.com\/calculus3\/wp-json\/wp\/v2\/types\/chapter"}],"author":[{"embeddable":true,"href":"https:\/\/courses.lumenlearning.com\/calculus3\/wp-json\/wp\/v2\/users\/349141"}],"version-history":[{"count":1,"href":"https:\/\/courses.lumenlearning.com\/calculus3\/wp-json\/pressbooks\/v2\/chapters\/1161\/revisions"}],"predecessor-version":[{"id":4698,"href":"https:\/\/courses.lumenlearning.com\/calculus3\/wp-json\/pressbooks\/v2\/chapters\/1161\/revisions\/4698"}],"part":[{"href":"https:\/\/courses.lumenlearning.com\/calculus3\/wp-json\/pressbooks\/v2\/parts\/1150"}],"metadata":[{"href":"https:\/\/courses.lumenlearning.com\/calculus3\/wp-json\/pressbooks\/v2\/chapters\/1161\/metadata\/"}],"wp:attachment":[{"href":"https:\/\/courses.lumenlearning.com\/calculus3\/wp-json\/wp\/v2\/media?parent=1161"}],"wp:term":[{"taxonomy":"chapter-type","embeddable":true,"href":"https:\/\/courses.lumenlearning.com\/calculus3\/wp-json\/pressbooks\/v2\/chapter-type?post=1161"},{"taxonomy":"contributor","embeddable":true,"href":"https:\/\/courses.lumenlearning.com\/calculus3\/wp-json\/wp\/v2\/contributor?post=1161"},{"taxonomy":"license","embeddable":true,"href":"https:\/\/courses.lumenlearning.com\/calculus3\/wp-json\/wp\/v2\/license?post=1161"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}